 # Editorial request palindromic numbers

would someone be kind enough to explain the solution for this problem (or the idea of it)in the answers or give an edirotial thanks!
Edit : I admit I am not very good with numbers

2 Likes

We can solve this problem by taking bases from 1 to N and see if our N in each of these Bases gives palindrome.On first occurrence of such base (which will be the smallest one), we break from loop and print the answer .But this will take a bit longer time especially when n can be 10^10.

So, how will this above algorithm be ?? (we will keep on optimizing it till we obtain final algorithm)

``````for Base =2 to N do
if(ispalidrome(N,Base)==True)
print the first such Base and exit the loop
``````

Now how do we define the method ispalindrome(N,base) ??

we see let N in base b , be written as N= a4b^4 + a3b^3 +a2b^2 +a1b^1 +a0b^0 .
This number N will be palindrome if we reverse the ai’s and yet the number is same i.e.
N= a0
b^4 + a1b^3 +a2b^2 +a3b^1 +a4b^0 .This is because if it is palindrome in base b then Decimal representation of reverse form should also be N.

We will represent ‘%’ as remainder operator

``````ispalindrome(N,base)
{
temp=num;
reverse=0;

while(temp!=0)
{
reverse =reverse *base + temp%base;
temp=temp/base;

}

if(reverse is equal to num)
return true;
else
return false;

}
``````

Just dry run it and see it.

Now our algorithm is done but we need to optimize it more. We make use of a following property.

Let a be square root of N. Then all bases greater than ‘a’ , will represent N in only 2 bits. (we will prove it later but for the time being use it.)

Th property says like this , say 10 has to represented in base 2. So 10 =1010 .
Now square root of 10 is approx 3 in integers. so let’s take base 4,5,…,9,10 .They will need at max 2 bits to represent 10. 10 in base 5 = 11 ,in base 6 =14 etc…

So how do we use it ? so we loop in 2 parts .

``````for Base =2 to square root of N do
if(ispalidrome(N,Base)==True)
print the first such Base and exit the loop
for Base = square root of N+1 to N
if the last 2 digits are equal
print the first such Base and exit the loop
``````

How do we check the last two digits ? n%(base) and (n/base) are the last 2 digits

But This algorithm will also take much time . Let’s optimize it even more .
The loop “for Base = square root of N+1 to N” is very long especially when n= 10^10.
We apply following trick.

let N= dd in some base ‘b’.

‘dd’ is the last 2 digits and base ‘b’ is a base greater than square root of N. Clearly such condition will arise when we are unable to find some base <=square root of N , where N is palindrome. Now we know it will consist of only 2 digits and for palindrome those digits have to equal.

So, N=d*b +d (write in decimal notation)
N=(b+1)*d => b= N/d-1

So instead of scanning all bases from square root of N+1 to N , we scan only bases of form N/d -1 . clearly , b+1 =n/d , so if LHS is integer RHS has to be integer. So, d must divide N. Also, b >square root of N. let the square root be ‘sqroot’ . so N/d-1 >sqroot => d< N/(sqroot+1) . So , d will go from N/(sqroot+1) to 1 . For each d , base = (N/d)-1 .Also , d being digits in base ‘b’, d<base.

``````for Base =2 to square root of N do
if(ispalidrome(N,Base)==True)
print the first such Base and exit the loop
for(d=N/(sqroot+1);d>=1;--d)
base =N/d-1;
if(N%d==0 && d<base && N/base == N%base )
Print the base which is (N/d)-1 and exit the loop
``````

Here is a c-code to implement the ALGORITHM .Here, take correct data type to accommodate numbers of order 10^10.

Note 1,2 are exceptional cases that cannot be dealt with the following algorithm . You have to explicitly mention their bases . for 1 it is 2 . For 2 it is 3 .(Single bits are palindrome.)

Now come to our question : Why all bases greater than square root of N , will represent N in only 2 bits.

Let a^2 <= N <(a+1)^2

Now square root of N is a. so we claim a+1,a+2,a+3,… will require at max 2 bits to represent N.

We note that N<(a+1)^2 .let’s convert both sides in base (a+1). so N-representation < 100. clearly this means N representation is of 2 bits. For bases higher than a+1 , it will be even more compact.
Hence proved.

Exceptions are 1 and 2 as we can see that by definition, bases are positive integers >=2 if N=1 or 2 ,then a is not a valid base .

Hope this explanation helps.

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this is very wonderful and complete explanation thank you good sir.

Thanks a lot @empty_life. You made me learn something. I owe this one to you